w
w
ap
eP
m
e
tr
.X
w
CAMBRIDGE INTERNATIONAL EXAMINATIONS
9691 COMPUTING
9691/21
Paper 2 (Written Paper), maximum raw mark 75
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
om
.c
MARK SCHEME for the May/June 2014 series
s
er
GCE Advanced Subsidiary Level and GCE Advanced Level
Page 2
1
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
(a) (i) Mark as follows:
1 mark for heading/introduction
1 mark for method of displaying numbers 1 – 10
e.g. dropdown box/radio buttons/grid of numbers)
1 mark for method to move to next screen (ignore exit/cancel) (accept next/ok/enter)
e.g. button // a label telling child what to do
[3]
(ii) 1 mark for explanation that fits design of (a)(i)
e.g. clicking on/touching/pressing button/box/number/icon
[1]
(b) Mark as follows:
• 1 mark for each box correctly translated into chosen programming language
• Identifiers must be the same as given in flowchart
• Give 1 mark for loop header and end correctly coded (must be a FOR loop)
• Ignore any declarations
• If candidate only says “Visual Basic” (no version number) use the mark scheme that best
fits the answer
• If language given is “Pseudocode”, give no marks
[max 6]
VB6 – accept console mode answers
Number = InputBox("")
Msg = Number & " Times Table" & vbCrLf
For i = 1 To 10
Result = i * Number
Msg = Msg & i & "x" & Number & "=" & Result & vbCrLf
Next i
Msg = Msg & "Press any key"
MsgBox(Msg)
VB.NET/VB 2005 etc.
Number = Console.ReadLine();
Console.WriteLine(Number & " Times Table");
For i = 1 To 10
Result = i * Number
Console.WriteLine(i & "x" & Number & "=" & Result)
Next i
Console.WriteLine("Press any key")
© Cambridge International Examinations 2014
Page 3
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
QBASIC
INPUT Number;
PRINT Number; " Times Table"
FOR i = 1 TO 10
Result = i * Number
PRINT i; "x"; Number; "="; Result
NEXT i
PRINT "Press any key"
PASCAL
ReadLn(Number);
WriteLn(Number, ' Times Table');
For i := 1 To 10 Do
Begin
Result := i * Number;
WriteLn(i, 'x', Number, '=', Result); // semicolon optional here
End;
// Begin End are part of the loop
structure
WriteLn('Press any key');
PYTHON
Number = int(input())
print(Number, "Times Table")
for i in range(1,11) :
Result = i * Number
print(i, "x", Number, "=", Result)
print("Press any key")
(c) PROCEDURE ShowMultiplicationGrid (Number1, Number2)
1 mark if Number1 and
FOR Row 1 TO Number1
// 1 mark
Number2 switched
round
FOR Column 1 TO Number2
// 1 mark
OUTPUT '*'
// 1 mark – accept '*' or "*"
ENDFOR
// 1 mark – accept NEXT column
OUTPUT NewLine
// 1 mark – must be in outer loop
ENDFOR
ENDPROCEDURE
[5]
© Cambridge International Examinations 2014
Page 4
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
(d) 1 mark for each correctly labelled shape (accept full text instead of number)
FUNCTION CheckAnswerCorrect(Number1, Number2)
R ← Number1 * Number 2
INPUT
AnswerString
No
Yes
Convert AnswerString to
AnswerNumber
No
No
Yes
Yes
Return TRUE
ENDFUNCTION
[10]
© Cambridge International Examinations 2014
Page 5
Mark Scheme
GCE AS/A LEVEL – May/June 2014
(2) (a) FUNCTION TestScoreTotal RETURNS INTEGER
DECLARE AnswerCorrect, Finish : BOOLEAN
DECLARE Number1, Number2, Score : INTEGER
Score 0
Syllabus
9691
Paper
21
// 1 mark
// 1 mark
// 1 mark
// 1 mark
Finish FALSE
REPEAT
Number1 Random(10)
// accept X instead of 10
Number2 Random(10)
// 1 mark
Display(Number1, Number2)
AnswerCorrect CheckAnswerCorrect(Number1, Number2)
IF AnswerCorrect = TRUE
// 1 mark
THEN Score Score + 1
ELSE Finish TRUE
// 1 mark
ENDIF
// accept =
instead of UNTIL Finish = TRUE
// 1 mark
RETURN Score
// 1 mark
ENDFUNCTION
[9]
•
•
•
If candidate only says “Visual Basic” (no version number) use the mark scheme that best
fits the answer
If language given is “Pseudocode”, give no marks
Identifiers must be as given in the question
Mark as follows:
(b) (i) 2 marks
• Correct dimension 1 mark
• Correct data types 1 mark
[2]
(ii) 2 marks
• 1 mark for structure
• 1 mark for correct fields (Name, BestScore) listed
[2]
(iii) 2 marks
• 1 mark for correct name (Student) and size
• 1 mark for correct record type (StudentScore)
[2]
(iv) 3 marks
• 1 mark for correctly addressed array elements (Student(3))
• 1 mark for correct reference to individual fields (.Name, .BestScore)
• 1 mark for correct assignment of values
[3]
© Cambridge International Examinations 2014
Page 6
Mark Scheme
GCE AS/A LEVEL – May/June 2014
VB6
QUESTION 2(b)(i)
Dim Name(30) As String
Dim BestScore(30) As Integer
QUESTION 2(b)(ii)
Type StudentScore
Name As String
BestScore As Integer
End Type
QUESTION 2(b)(iii)
Dim Student(30) As StudentScore
QUESTION 2(b)(iv)
Student(3).Name = "Anji"
Student(3).BestScore = 15
VB.NET/VB 2005 etc.
QUESTION 2(b)(i)
Dim Name(30) As String
Dim BestScore(30) As Integer
QUESTION 2(b)(ii)
Structure StudentScore
Dim Name As String
Dim BestScore As Integer
End Structure
QUESTION 2(b)(iii)
Dim Student(30) As StudentScore
QUESTION 2(b)(iv)
Student(3).Name = "Anji"
Student(3).BestScore = 15
QBASIC
QUESTION 2(b)(i)
DIM Name(30) AS STRING * 15
DIM BestScore(30) AS INTEGER
© Cambridge International Examinations 2014
Syllabus
9691
Paper
21
Page 7
Mark Scheme
GCE AS/A LEVEL – May/June 2014
QUESTION 2(b)(ii)
TYPE StudentScore
Name AS STRING * 15
BestScore AS INTEGER
END TYPE
QUESTION 2(b)(iii)
DIM Student(30) AS StudentScore
QUESTION 2(b)(iv)
Student(3).Name = "Anji"
Student(3).BestScore = 15
PASCAL
QUESTION 2(b)(i)
Name : Array[1..30] Of String;
BestScore : Array[1..30] Of Integer;
QUESTION 2(b)(ii)
Type StudentScore = Record
Name : String[15];
BestScore : Integer;
End;
QUESTION 2(b)(iii)
Student : Array[1..30] Of StudentScore;
QUESTION 2(b)(iv)
Student[3].Name := 'Anji';
Student[3].BestScore := 15;
© Cambridge International Examinations 2014
Syllabus
9691
Paper
21
Page 8
Mark Scheme
GCE AS/A LEVEL – May/June 2014
PYTHON
QUESTION 2(b)(i)
Name = ["" for i in range(30)]
BestScore = [0 for i in range(30)]
or
Name = []
BestScore = []
for i in range(30) :
Name.append("")
BestScore.append(0)
QUESTION 2(b)(ii)
class StudentScore :
def __init__(self) :
Name = ""
BestScore = 0
QUESTION 2(b)(iii)
Student = [StudentScore() for i in range(30)]
or
Student = []
for i in range(30) :
Student.append(StudentScore())
QUESTION 2(b)(iv)
Student[2].Name = "Anji"
Student[2].BestScore = 15
Mark as follows:
•
•
•
•
•
•
Procedure/sub SaveToFile
Open (for writing) StudentFile
Initialise index value correctly
Loop through student array
Write record to file StudentFile
Close file StudentFile
© Cambridge International Examinations 2014
Syllabus
9691
Paper
21
Page 9
(c)
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
VB6
Sub SaveToFile()
Dim i as Integer
Open "StudentFile" For Output As #1
For i = 1 To 30
Write#1, Student(i).Name, Student(i).BestScore
Next i
Close#1
End Sub
VB.NET/VB 2005 etc.
Sub SaveToFile()
Dim Writer As BinaryWriter
Writer
=
new
BinaryWriter(File.Open("StudentFile",
FileMode.Create))
For i = 1 To 30
Writer.Write(Student(i).Name)
Writer.Write(Student(i).BestScore)
Next i
Writer.Close()
End Sub
QBASIC
SUB SaveToFile(Student() AS StudentScore)
OPEN "StudentFile" FOR OUTPUT AS #1
FOR i = 1 TO 30
WRITE #1, Student(i).Name, Student(i).BestScore
NEXT i
CLOSE #1
END SUB
PASCAL
Var Students : File Of StudentScore;
Procedure SaveToFile;
Var i : Integer;
Begin
Assign(Students, 'StudentFile');
Rewrite(Students);
For i := 1 To 30 Do
Write(Students, Student[i]);
Close(Students);
End;
© Cambridge International Examinations 2014
Page 10
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
PYTHON
import pickle
def SaveToFile() :
Students = open("StudentFile", "wb")
for i in range(30) :
pickle.dump(Student[i], Students)
Students.close()
or (as a text file)
def SaveToFile() :
Students = open("StudentFile", "w")
for i in range(30) :
Students.write(Student[i].Name + "\n")
Students.write(str(Student[i].BestScore) + "\n")
Students.close()
[6]
© Cambridge International Examinations 2014
Page 11
3
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
(a) (i) FUNCTION FindArrayIndex(ChildName : STRING) RETURNS INTEGER
i 0
REPEAT
i i + 1
UNTIL Student[i].Name = ChildName
RETURN i
Another method:
FUNCTION FindArrayIndex(ChildName : STRING) RETURNS INTEGER
Found FALSE
i 1
WHILE NOT (Found = TRUE)
IF Student[i].Name = ChildName
THEN Found TRUE
ELSE i i + 1
ENDIF
ENDWHILE
RETURN i
[max 5]
(ii) suggestion of checking for end of array
// checking whole array and setting flag if not found
Special value/error code/number returned if name doesn’t exist (e.g. –1)
[2]
(b) e.g.
– turn all characters into lower/upper case
– before saving/searching // on input
[2]
© Cambridge International Examinations 2014
Page 12
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
(4) (i) – it calls itself
[1]
(ii)
Call
Function
Number call
s
x
RIGHT
(s,x - 1)
1
Y('BYTE')
'BYTE' 4
'YTE'
2
Y('YTE')
'YTE'
3
'TE'
3
Y('TE')
'TE'
2
'E'
4
Y('E')
'E'
1
LEFT(s,1)
Return
value
'E'
(3)
'T'
'ET'
(2)
'Y'
'ETY'
(1)
'B'
'ETYB'
1 mark per correct column
(iii) reverses the string
[7]
[1]
(iv) – indentation
comment/annotation/remarks
(v) – identifiers not meaningful/sensible // identifiers are just single characters
(vi) if answer is recursive, no marks
There are many different ways of solving this. The following are examples:
n '' /empty string
REPEAT
x LENGTH(s)
z LEFT(s,1)
s RIGHT(s, x - 1)
n z + n
// y concat(z,n)
UNTIL s = ''
RETURN n
© Cambridge International Examinations 2014
[2]
[1]
Page 13
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
21
Or
n '' /empty string
x LENGTH(s)
FOR i 1 TO x
z LEFT(s,1)
s RIGHT(s, x - i)
n z + n
// n concat(z,n)
ENDFOR
RETURN n
n '' /empty string
x LENGTH(s)
FOR i 1 TO x
z MID(s,i,1)
n z + n
ENDFOR
RETURN n
// n concat(z,n)
Mark as follows:
– start with empty string
– correct loop structure
– correct loop count/termination
– pick single character (from string s) consecutively
– concatenate single character to correct end of new string
– return newly formed string
[5]
© Cambridge International Examinations 2014